1940 United States presidential election in Rhode Island

The 1940 United States presidential election in Rhode Island took place on November 5, 1940. All contemporary 48 states were part of the 1940 United States presidential election. State voters chose 4 electors to the Electoral College, which selected the president and vice president.

Main article: 1940 United States presidential election

1940 United States presidential election in Rhode Island

November 5, 1940

Nominee Franklin D. Roosevelt Wendell Willkie Party Democratic Republican Home state New York Indiana Running mate Henry A. Wallace Charles L. McNary Electoral vote 4 0 Popular vote 182,182 138,653 Percentage 56.73% 43.17%

County Results Roosevelt   50-60% Willkie   50-60%   60-70%

President before election Franklin D. Roosevelt Democratic Elected President Franklin D. Roosevelt Democratic

Rhode Island was won by incumbent Democratic President Franklin D. Roosevelt of New York, who was running against Republican businessman Wendell Willkie of Indiana. Roosevelt ran with Henry A. Wallace of Iowa as his running mate, and Willkie ran with Senator Charles L. McNary of Oregon.

Roosevelt won Rhode Island by a margin of 13.56%.

Results

1940 United States presidential election in Rhode Island[1]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Democratic	Franklin Delano Roosevelt of New York	Henry Agard Wallace of Iowa	182,182	56.73%	4	100.00%
	Republican	Wendell Willkie of Indiana	Charles Linza McNary of Oregon	138,653	43.27%	0	0.00%
	Communist	Earl Russell Browder of Kansas	James W. Ford of New York	239	0.07%	0	0.00%
	Prohibition	Roger Ward Babson of Massachusetts	Edgar Moorman of Illinois	74	0.02%	0	0.00%
Total				321,148	100.00%	4	100.00%

References

1. "1940 Presidential General Election Results – Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.